Here is a solution for $n=2d^2-2$ with $|S|=4d-3.$ It is just Gerry Myerson's scheme (1 2 3 4 5) with a very slight adjustment, so the same asymptotics.
$$0,1,\dotsc, d-1$$ then $$2d-1,3d-1,4d-1,\dotsc ,(2d-2)d-1$$ then $$n-d+1,n-d+2,\dotsc,n.$$
Note that $n-d+1=(2d-1)d-1$.
Here are some of Rob Pratt's reported values (which I have no reason to question) with $n=2d^2-2$ in bold.
$$\begin{array}{c|c} \lvert S\rvert & n \\ \hline 5 & \text{5, }\mathbf6 \\ 9 & \text{14, 15, }\mathbf{16} \\ 13 & \text{28, 29, }\mathbf{30}\text{, 32} \\ 14 & \text{31, 33, 34, 35, 36} \\ 17 & \text{47, }\mathbf{48}\text{, 49, 50, and perhaps a few larger $n$?} \end{array}$$ It would look nicer if $n=31$ required $\lvert S\rvert=14$; this is the only case reported with $\lvert S\rvert=m(n) \lt m(n-1)$.
I would be curious to see the optimal sets $\lvert S\rvert$.
